Problem: $\dfrac{ s + 3t }{ 4 } = \dfrac{ -10s + 10u }{ 7 }$ Solve for $s$.
Solution: Multiply both sides by the left denominator. $\dfrac{ s + 3t }{ {4} } = \dfrac{ -10s + 10u }{ 7 }$ ${4} \cdot \dfrac{ s + 3t }{ {4} } = {4} \cdot \dfrac{ -10s + 10u }{ 7 }$ $s + 3t = {4} \cdot \dfrac { -10s + 10u }{ 7 }$ Multiply both sides by the right denominator. $s + 3t = 4 \cdot \dfrac{ -10s + 10u }{ {7} }$ ${7} \cdot \left( s + 3t \right) = {7} \cdot 4 \cdot \dfrac{ -10s + 10u }{ {7} }$ ${7} \cdot \left( s + 3t \right) = 4 \cdot \left( -10s + 10u \right)$ Distribute both sides ${7} \cdot \left( s + 3t \right) = {4} \cdot \left( -10s + 10u \right)$ ${7}s + {21}t = -{40}s + {40}u$ Combine $s$ terms on the left. ${7s} + 21t = -{40s} + 40u$ ${47s} + 21t = 40u$ Move the $t$ term to the right. $47s + {21t} = 40u$ $47s = 40u - {21t}$ Isolate $s$ by dividing both sides by its coefficient. ${47}s = 40u - 21t$ $s = \dfrac{ 40u - 21t }{ {47} }$